To solve this problem, we need to understand the relationship between magnification $(m), \\$the type of mirror, and the nature of the image.$\\$Magnification $(m)$ is defined as the ratio of the height of the image to the height of the object.$\\$It is given by $m = \frac{-v}{u},$ where $v$ is the image distance and $u$ is the object distance.$\\$Let's analyze each case:$\\$(A) $m = -2: \\$• The negative sign indicates the image is real and inverted.$\\$• A concave mirror can produce a real image with magnification greater than 1.$\\$• Therefore, it corresponds to a concave mirror and a real image.$\\$(B) $m = -\frac{1}{2}: \\$• The negative sign indicates the image is real and inverted.$\\$• A concave mirror can produce a real image with magnification less than 1.$\\$• Therefore, it corresponds to a concave mirror and a real image.$\\$(C) $m = +2: \\$• The positive sign indicates the image is virtual and erect.$\\$• A concave mirror can produce a virtual image with magnification greater than 1.$\\$• Therefore, it corresponds to a concave mirror and a virtual image.$\\$(D) $m = +\frac{1}{2}: \\$• The positive sign indicates the image is virtual and erect.$\\$• A convex mirror always produces a virtual image with magnification less than 1.$\\$• Therefore, it corresponds to a convex mirror and a virtual image.$\\$Matching the entries:$\\$(A) → (b) and (c); (B) → (b) and (c); (C) → (b) and (d); (D) → (a) and (d)$\\$This corresponds to Option 3.